The O(1) space constraint is where I got stuck.
Clarify the problem constraints and walk through a simple example to demonstrate the distribution process. Then, propose a single-pass algorithm that processes the string character by character, maintaining the current term and accumulated result, leveraging the fact that '+' only appears inside parentheses to avoid needing a stack. Finally, discuss how to achieve O(n) time and O(1) extra space by carefully managing state variables.
Pro tip: Emphasize that the restriction of '+' only inside parentheses allows a stackless solution; explicitly state that you are exploiting this constraint to avoid the typical stack-based approach for nested expressions.
Confirm that variables are single lowercase letters, '+' only appears inside parentheses, and the expression is fully parenthesized. Discuss edge cases like empty string, single variable, or nested parentheses.
Take a small expression like 'a*(b+c)' and manually distribute to show the expected output 'a*b + a*c'. This demonstrates understanding and sets the stage for the algorithm.
Propose a single left-to-right scan. Maintain a current term (string) and a result (string). When encountering '(', start a new term; when encountering ')', append the current term to the result; when encountering '*', multiply the current term by the next variable; when encountering a variable, append it to the current term. Since '+' only appears inside parentheses, we can handle it by splitting the current term into multiple terms and appending each to the result.
Explain that each character is processed once, so time is O(n). Extra space is O(1) beyond the output because we only use a few pointers and temporary strings that are part of the output.
Mention that a stack-based approach would be more general but uses O(n) space. Highlight that the stackless solution is possible due to the problem's constraints, showing awareness of trade-offs.
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This part felt more natural to me since using a stack for nested structure is pretty standard.
Clarify the problem: evaluate an expression with + and * operators, parentheses, and possibly nested parentheses, using a stack to handle multiplication segments. Outline a two-pass or single-pass stack-based algorithm that maintains a current segment and a result, pushing/popping on parentheses. Then analyze time and space complexity and discuss edge cases like negative numbers, multiple digits, and empty input.
Pro tip: Emphasize that the stack stores intermediate results and signs, and that multiplication has higher precedence, so you process it immediately while deferring addition. Mention that this approach naturally extends to nested parentheses by saving the current state before entering a parenthesis and restoring it after.
Confirm the operators (+, *), parentheses, and that numbers can be multi-digit. Ask about negative numbers, spaces, and whether division is included. State that you'll aim for O(n) time and O(n) space.
Use a stack to store the result so far and the sign before a parenthesis. Maintain a current number and a current multiplication segment. When encountering '+', add the segment to the result and start a new segment. When encountering '(', push the current result and sign, then reset. When encountering ')', finalize the current segment, apply the sign, add to the result, then pop and combine with the previous result.
Trace the algorithm on a nested expression like '2*(3+(4*5+6))' to show how the stack and segments are updated. Highlight how multiplication is handled immediately and addition is deferred.
State that each character is processed once, so time is O(n). Space is O(n) in the worst case due to stack depth for nested parentheses. Discuss edge cases: empty string, single number, negative numbers, multiple digits, and unbalanced parentheses.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.